Chemistry · Ch 1 — Basic Concepts of Chemistry and Chemical Calculations
Equivalent Mass of Acids, Bases, Salts, Oxidising Agents and Reducing Agents
Equivalent Mass of Acids, Bases, Salts, Oxidising Agents and Reducing Agents
The equivalence factor n -- and hence the equivalent mass formula E = (molar mass) ÷ n -- takes a different, specific meaning depending on the type of chemical entity:
- Acids: n = the basicity of the acid, i.e. the number of moles of ionisable H⁺ ions present in 1 mole of the acid. Example: H₂SO₄ has basicity 2 eq mol⁻¹ (two ionisable H⁺ per molecule) and molar mass (2×1) + 32 + (4×16) = 98 g mol⁻¹, so its gram equivalent mass = 98 ÷ 2 = 49 g eq⁻¹.
- Bases: n = the acidity of the base, i.e. the number of moles of ionisable OH⁻ ions present in 1 mole of the base. Example: KOH has acidity 1 eq mol⁻¹ and molar mass 39 + 16 + 1 = 56 g mol⁻¹, so its gram equivalent mass = 56 ÷ 1 = 56 g eq⁻¹.
- Oxidising or reducing agents: n = the number of moles of electrons gained (for an oxidising agent) or lost (for a reducing agent) by 1 mole of the reagent during the redox reaction. Example: KMnO₄ (molar mass 39 + 55 + (4×16) = 158 g mol⁻¹) acts as an oxidising agent in acid medium via MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so n = 5 eq mol⁻¹ and its gram equivalent mass = 158 ÷ 5 = 31.6 g eq⁻¹.
Why bother with a second concept alongside the mole? The mole concept needs a fully balanced chemical equation before you can work out how much of one reactant reacts with another. The gram equivalent concept does not -- by definition, equivalent amounts always react in a 1 : 1 ratio with each other, whatever the actual reaction is. This is exactly why the equivalent-mass route is the natural tool for redox reactions, where writing and balancing the full equation can be laborious, while the mole concept remains the natural tool for non-redox reactions. …
| Chemical entity | Equivalent factor (n) | Formula for equivalent mass (E) | Worked example |
|---|---|---|---|
| Acids | basicity (no. of moles of ionisable H⁺ in 1 mole of the acid) | E = molar mass of the acid ÷ basicity | H₂SO₄: basicity = 2 eq mol⁻¹, molar mass = 98 g mol⁻¹, gram equivalent mass = 98/2 = 49 g eq⁻¹ |
| Bases | acidity (no. of moles of ionisable OH⁻ in 1 mole of the base) | E = molar mass of the base ÷ acidity | KOH: acidity = 1 eq mol⁻¹, molar mass = 56 g mol⁻¹, gram equivalent mass = 56/1 = 56 g eq⁻¹ |
Worked out. Explains that the mole concept needs a balanced chemical equation to find reacting amounts, while the gram equivalent concept does not -- so, knowing the equivalent masses of KMnO₄ (31.6 g eq⁻¹) and anhydrous FeSO₄ (152 g eq⁻¹), one can say directly that 31.6 g of KMnO₄ reacts with 152 g of FeSO₄. The same result is then verified using the balanced equation 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O: 2 moles (316 g) of KMnO₄ react with 10 moles (1520 g) of FeSO₄, so 31.6 g of KMnO₄ reacts with (1520/316) × 31.6 = 152 g of FeSO₄ -- the same answer, confirming that 'we prefer mole concept for non-redox reactions and gram equivalent concep …
Worked out. An in-text practice box with two parts: (4a) 0.456 g of a metal gives 0.606 g of its chloride -- calculate the equivalent mass of the metal; (4b) calculate the equivalent mass of potassium dichromate given the reduction half-reaction in acid medium, Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. …